To solve the problem of maximizing the number of cakes that can be made from given quantities of flour and fat, we will formulate it as a linear programming problem and solve it graphically.
### Step 1: Define the Variables
Let:
- \( x \) = number of cakes of the first kind
- \( y \) = number of cakes of the second kind
### Step 2: Formulate the Constraints
From the problem, we know:
- The first kind of cake requires 200 g of flour and 25 g of fat.
- The second kind of cake requires 100 g of flour and 50 g of fat.
Given:
- Total flour available = 5 kg = 5000 g
- Total fat available = 1 kg = 1000 g
The constraints based on the ingredients are:
1. For flour:
\[
200x + 100y \leq 5000
\]
Dividing the entire equation by 100 gives:
\[
2x + y \leq 50 \quad \text{(Constraint 1)}
\]
2. For fat:
\[
25x + 50y \leq 1000
\]
Dividing the entire equation by 25 gives:
\[
x + 2y \leq 40 \quad \text{(Constraint 2)}
\]
### Step 3: Non-negativity Constraints
Since the number of cakes cannot be negative, we have:
\[
x \geq 0 \quad \text{and} \quad y \geq 0 \quad \text{(Constraint 3)}
\]
### Step 4: Objective Function
We want to maximize the total number of cakes:
\[
Z = x + y
\]
### Step 5: Graph the Constraints
1. **Graph Constraint 1**: \( 2x + y = 50 \)
- When \( x = 0 \), \( y = 50 \) (point A: (0, 50))
- When \( y = 0 \), \( x = 25 \) (point B: (25, 0))
2. **Graph Constraint 2**: \( x + 2y = 40 \)
- When \( x = 0 \), \( y = 20 \) (point C: (0, 20))
- When \( y = 0 \), \( x = 40 \) (point D: (40, 0))
### Step 6: Identify the Feasible Region
The feasible region is the area where all constraints overlap, which is bounded by the lines from the constraints and the axes.
### Step 7: Find the Corner Points
The corner points of the feasible region can be found by solving the equations of the lines:
1. Intersection of \( 2x + y = 50 \) and \( x + 2y = 40 \):
- From \( 2x + y = 50 \), we can express \( y = 50 - 2x \).
- Substitute into the second equation:
\[
x + 2(50 - 2x) = 40 \implies x + 100 - 4x = 40 \implies -3x = -60 \implies x = 20
\]
Substitute \( x = 20 \) back to find \( y \):
\[
y = 50 - 2(20) = 10 \quad \text{(Point E: (20, 10))}
\]
2. Other corner points are:
- Point A: (0, 20)
- Point B: (25, 0)
- Point D: (0, 0)
### Step 8: Evaluate the Objective Function at Each Corner Point
1. At (0, 20): \( Z = 0 + 20 = 20 \)
2. At (20, 10): \( Z = 20 + 10 = 30 \)
3. At (25, 0): \( Z = 25 + 0 = 25 \)
4. At (0, 0): \( Z = 0 + 0 = 0 \)
### Step 9: Determine the Maximum Value
The maximum value of \( Z \) occurs at the point (20, 10) where \( Z = 30 \).
### Conclusion
The maximum number of cakes that can be made is:
- 20 cakes of the first kind
- 10 cakes of the second kind
